# Smooth chart in what sense?

I have a question concerning smooth manifolds. As far as I've understand a smooth manifold is a pair of a manifold and a smooth atlas. Where smooth atlas means that the transition functions defiened on overlapping charts are smooth from $\mathbb{R}^n$ to $\mathbb{R}^n$. In particular a chart in a smooth atlas is (in some sense) only a homeomorphism. Still it seems like the smooth charts are in them selves to be smooth?

My concern relates to Lee J.M Introduction to smooth manifolds, where he defines the smooth atlas $x\rightarrow x^3$ from $\mathbb{R}$ to $\mathbb{R}$. and states that it is not smoothly equivalent to the identity. I.e this is an example of a smooth chart that is not given by a smoothly invertible function (in the usual sense). This makes me confused, since in the same book it is stated that the coordinate charts are local diffeomorphisms between $\mathbb{R}^n$ and some open nbh of the manifold.

When we say that the coordinate charts locally gives a diffeomorphism do we mean this with respect to the very same coordinate functions, i.e we are saying that a chart is a diffeomorphism because composing it with its inverse gives us the identity, which is a diffeomorphism?

Any comments or clarifications are welcome!

Let $(\phi, U)$ be a coordinate chart on the manifold $M$. Suppose that $V \subset \mathbb R^n$ is the image of $\phi$. Observe that an open subset of a smooth manifold is still a smooth manifold. Thus $U$ and $V$ are smooth manifolds, with smooth atlases coming respectively from $M$ and $\mathbb R^n.$ With these definitions in place, let us show that $\phi$ is a diffeomorphism from $U$ to $V$. First, it is a homeomorphism, from the definition of coordinate chart. Let us show that $\phi$ is smooth. Note that the identity map $id: V \to V$ is a coordinate chart on the smooth manifold $V$. Writing $\phi$ with respect to the coordinate charts $(id, V)$ and $(\phi, U)$ we get that, in local coordinates on $V$ and $U,$ $$id \circ \phi \circ \phi^{-1} = id,$$ in particular, $\phi$ when written in local coordinates is a diffeomorphism.
So the short answer is: Yes, in the definition of a manifold, the coordinate chart is a homeomorphism, not a diffeomorphism. But if you regard the image of a coordinate chart as a manifold, and then use the sneaky trick of writing $\phi$ in local coordinates w.r.t. itself, you see that $\phi$ satisfies the definition of a diffeomorphism.
Let $\mathcal A$ be the maximal atlas containing the identity map $id:\mathbb R \to \mathbb R$. Let $\mathcal B$ be the maximal atlas containing the chart $x \mapsto x^3.$ Then $\mathcal A \neq \mathcal B$, so that the smooth manifolds $(\mathbb R, \mathcal A)$ and $(\mathbb R, \mathcal B)$ are not the same. Nevertheless, they are diffeomorphic. The mapping $f:(\mathbb R, \mathcal A) \to (\mathbb R, \mathcal B)$ given by $f(x) = x^3$ is a diffeomorphism between them.
• Note that $f$ should read $f(x) = x^{1/3}$, since you want to compose it with the chart $\phi : x \mapsto x^3$ from $\mathcal{B}$, i.e. $\phi \circ f \circ id_\mathbb{R}^{-1}$. – el_tenedor Mar 28 '16 at 14:19